$\large \sqrt[3] {a} -\sqrt[3] {b}- \sqrt[3] {c} =16$

$\large \sqrt[4] {a} -\sqrt[4] {b} - \sqrt[4]{c}=8$

$\sqrt[6] a -\sqrt[6] b - \sqrt[6] c =4$

I was able to find one solution:

$\large (\sqrt [6]a, \sqrt [6]b, \sqrt [6] c)\rightarrow (x^2, -y^2, -z^2)$ respectively

$\large x^2+y^2+z^2=4$

$\large x^3+y^3+z^3=8$


Squaring the first equation,



Two of the variables have to be equal to $0$. Without loss of generality, let $\large (y, z)=0$ This means that $\large x=2$ , so the solution $\large (a, b, c)$ is $\large (4^6, 0, 0)$

Is there more that one solution? It really feels like there should be more since there are all these high powers.

  • $\begingroup$ I am weak on minus signs. Substituting letting $a=x^{12}$ and so on sounds like a good idea. But then we will get some minus signs on substituting. Of course the solution you found works, and the missing minus signs make no difference since $y=z=0$. $\endgroup$ – André Nicolas Jul 1 '13 at 6:32
  • $\begingroup$ @AndréNicolas I used the minus signs in order to make the equations with x, y, z to have only plus signs, but I just realized that it would only make it so for thee equation with the squares. I need to revise this a little bit. $\endgroup$ – Ovi Jul 1 '13 at 6:35
  • $\begingroup$ @Ovi If $$y^2=-\sqrt[6]{b}$$ how did you get $$y^3=-\sqrt[4]{b}$$ $\endgroup$ – Paracosmiste Jul 1 '13 at 7:39
  • $\begingroup$ Where did you take this problem from? $\endgroup$ – Paracosmiste Jul 1 '13 at 9:11
  • $\begingroup$ @metacompactness I realized that, but that can be fixed by making $\sqrt[3] a , \sqrt[3] b, \sqrt[3] c \rightarrow (x^4, -y^4, -z^4)$ . Then the third equation will be $x^4+y^4+z^4=16$ , and the first equation will be $x^2-y^2-z^2$ but when I square this I get a similar result and I can still make the substitution $x^4+y^4+z^4=16$ and be left with $0=yz-xy-xz$ . This not only allows for the solution that I already found, but since all the terms aren't positive anymore it also leaves room for $(y, z) \not = 0$ $\endgroup$ – Ovi Jul 1 '13 at 17:01

Let $x=\sqrt[12]{a}$, $y=\sqrt[12]{b}$ and $z=\sqrt[12]{c}$. The system becomes: \begin{equation} x^4-y^4-z^4=16~..................................(1)\\ x^3-y^3-z^3=8~.................................. .(2)\\ x^2-y^2-z^2=4~...................................(3) \end{equation} Squaring equation $(3)$ and adding it to equation $(1)$, we obtain: $$\left(x^2\right)^2-\left(y^2+z^2\right)x^2+y^2 z^2-16=0$$ whose discriminant is $\Delta=\left(y^2-z^2\right)^2+64$. The sum of its two solutions $x_1^2$ and $x_2^2$ is $x_1^2+x_2^2=y^2+z^2$.

Now, let us substitute $x_1$ and $x_2$ in equation $(3)$, we'll obtain: \begin{equation} x_1^2-y^2-z^2=4\\ x_2^2-y^2-z^2=4 \end{equation} Adding the last two equations gives us: $$y^2+z^2=-8$$ This last equation and equation $(3)$ gives us: $x^2=-4$ that is $x=\pm 2i$ Putting the values of $x$ in equation $(1)$, we obtain $$y^4+z^4=0~$$ which together with the equation $$\left(y^2+z^2\right)^2=(-8)^2$$ gives us: $y^2 z^2=32$

So we have two numbers $y^2$ and $z^2$ whose sum is $-8$ and whose product is $32$: they are the uolutions of the quadratic equation $u^2+8u+32=0$ so $$y=2^{\frac{5}{4}}\exp\left(\frac{3i\pi}{8}\right)$$ and $$y=2^{\frac{5}{4}}\exp\left(\frac{5i\pi}{8}\right)$$ or vice versa. So the solution I obtained until now are: $$(x,y,z)=\left(2i,2^{\frac{5}{4}}\exp\left(\frac{5i\pi}{8}\right),2^{\frac{5}{4}}\exp\left(\frac{3i\pi}{8}\right)\right)$$ $$(x,y,z)=\left(2i,2^{\frac{5}{4}}\exp\left(\frac{3i\pi}{8}\right),2^{\frac{5}{4}}\exp\left(\frac{5i\pi}{8}\right)\right)$$ $$(x,y,z)=\left(-2i,2^{\frac{5}{4}}\exp\left(\frac{5i\pi}{8}\right),2^{\frac{5}{4}}\exp\left(\frac{3i\pi}{8}\right)\right)$$ $$(x,y,z)=\left(-2i,2^{\frac{5}{4}}\exp\left(\frac{3i\pi}{8}\right),2^{\frac{5}{4}}\exp\left(\frac{5i\pi}{8}\right)\right)$$

I run out of time; I didn't use equation $(2)$ yet to find other solutions.

  • $\begingroup$ In the case of complex $y$ and $z$ the discriminant $\Delta$ is not necessarily positive. $\endgroup$ – user64494 Jul 1 '13 at 9:16
  • $\begingroup$ @user64494 You're right. $\endgroup$ – Paracosmiste Jul 1 '13 at 9:20
  • $\begingroup$ @ metacompacthess : Also $$(x,y,z)=\left(2i,2^{\frac{5}{4}}\exp\left(\frac{5i\pi}{8}\right),2^{\frac{5}{4}}\exp\left(\frac{3i\pi}{8}\right)\right) $$ does not satisfy the equation $x^3-y^3-z^3 = 8$. $\endgroup$ – user64494 Jul 1 '13 at 9:25

Maple also produces 6 complex solutions $\{x = -.8435457951+1.254365878i, y = 1.285003263-1.910820082i, z = -.7543658777-1.852277550i\}, \{x = -.8435457951+1.254365878i, y = -.7543658778-1.852277551i, z = 1.285003263-1.910820081i\}, \{x = -1.312908410, y = .9693626140+1.441457468i, z = .969362615-1.441457468i\}, \{x = -1.312908410, y = .9693626140-1.441457468i, z = .969362615+1.441457468i\}, \{x = -.8435457951-1.254365878i, y = 1.285003263+1.910820082i, z = -.7543658777+1.852277550i\}, \{x = -.8435457951-1.254365878i, y = -.7543658775+1.852277551i, z = 1.285003263+1.910820081i\}$ of the system $\{x^2-y^2-z^2 = 4, x^3-y^3-z^3 = 8, x^4-y^4-z^4 = 16\}.$ It is too long for a comment.

  • $\begingroup$ The solution is expressed in terms of the roots of the polynomial $2t^3+6t^2+9t+6.$ $\endgroup$ – user64494 Jul 1 '13 at 8:36
  • $\begingroup$ How is that? The polynomial is third degree so it only has three roots. $\endgroup$ – Ovi Jul 1 '13 at 17:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.